An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 CCEA A Level Technology and Design 8900 paper. Not affiliated with or reproduced from CCEA.
Section A: Electronic and Microelectronic Control Systems
Answer both Question 1 and Question 2 if choosing the Electronics pathway. Complete Question 1(f) and Question 2(d)(ii) on the dedicated answer pages.
16 Question · 80 marks
Question 1 · Short Answer & Calculation
3 marks
A voltage divider circuit consists of two resistors, \( R_1 = 4.7\ \text{k}\Omega \) and \( R_2 = 2.2\ \text{k}\Omega \), connected in series across a 9 V supply, with \( V_{out} \) taken across \( R_2 \). Calculate \( V_{out} \). Show your working.
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Worked solution
The voltage divider formula is \( V_{out} = V_{in} \times \frac{R_2}{R_1+R_2} \). Substituting the values: \( V_{out} = 9 \times \frac{2200}{4700+2200} = 9 \times \frac{2200}{6900} = 2.87\ \text{V} \) (to 3 s.f.).
An inverting operational amplifier circuit has an input resistor \( R_{in} = 10\ \text{k}\Omega \) and a feedback resistor \( R_f = 47\ \text{k}\Omega \). Calculate the voltage gain of the circuit, and state what the sign of your answer indicates about the output signal.
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Worked solution
The inverting amplifier gain formula is \( A_v = -\frac{R_f}{R_{in}} \). Substituting: \( A_v = -\frac{47}{10} = -4.7 \). The negative sign shows that the output signal is inverted relative to the input (a positive-going input produces a negative-going change at the output, and vice versa).
Marking scheme
1 mark: correct formula used \( \left(A_v = -\frac{R_f}{R_{in}}\right) \); 1 mark: \( A_v = -4.7 \); 1 mark: correctly explains the negative sign means the output is inverted. [3]
Question 3 · Short Answer & Calculation
3 marks
A non-inverting operational amplifier circuit has a feedback resistor \( R_f = 33\ \text{k}\Omega \) and a resistor to ground \( R_{in} = 10\ \text{k}\Omega \). Calculate the voltage gain of the circuit.
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Worked solution
The non-inverting amplifier gain formula is \( A_v = 1 + \frac{R_f}{R_{in}} \). Substituting: \( A_v = 1 + \frac{33}{10} = 1 + 3.3 = 4.3 \).
A Zener diode voltage regulator uses a 12 V Zener diode connected to an 18 V unregulated supply through a series resistor. The Zener current is to be limited to 15 mA. Calculate the value of the series resistor required.
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Worked solution
The series resistor formula for a Zener regulator is \( R = \frac{V_{in}-V_z}{I_z} \). Substituting: \( R = \frac{18-12}{0.015} = \frac{6}{0.015} = 400\ \Omega \).
Marking scheme
1 mark: correct formula used \( \left(R=\frac{V_{in}-V_z}{I_z}\right) \); 1 mark: correct substitution shown; 1 mark: \( R = 400\ \Omega \). [3]
Question 5 · Short Answer & Calculation
2 marks
State two safety precautions that should be followed when constructing and testing electronic circuits in a school or college laboratory.
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Worked solution
Safe practice when building and testing electronic circuits includes disconnecting the power supply before making or altering any circuit connections, to avoid short circuits or shock; checking the correct polarity of components such as electrolytic capacitors and diodes before applying power, since connecting these incorrectly can cause them to fail, overheat or (for electrolytic capacitors) rupture; not exceeding the rated voltage or current of any component; and wearing suitable eye protection when soldering.
Marking scheme
1 mark each for any two valid safety precautions, e.g. disconnect power before altering circuit; check polarity of components before powering up; do not exceed rated voltage/current; wear eye protection when soldering. [2]
Question 6 · Short Answer & Calculation
2 marks
State the property of a light dependent resistor (LDR) that makes it suitable for use as a light-sensing input component in a control system.
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Worked solution
An LDR's resistance changes with the amount of light falling on it: its resistance is high in darkness and falls as light intensity increases. Placed in a voltage divider circuit with a fixed resistor, this changing resistance produces a voltage that varies predictably with light level, allowing it to be used as a light-sensing input to a control system.
Marking scheme
1 mark: correctly states resistance decreases with increasing light (or increases in darkness); 1 mark: correctly links this to producing a varying voltage in a voltage divider circuit. [2]
Question 7 · Short Answer & Calculation
2 marks
Explain, using one example, the difference between on/off control and continuous control in an electronic control system.
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Worked solution
On/off control switches the output fully on or fully off depending on whether the input has crossed a set threshold; for example, a heater controlled this way switches fully on when the temperature falls below a set point and fully off once it rises above it. Continuous (proportional) control instead varies the output smoothly in proportion to the changing input signal; for example, a dimmer circuit smoothly varies an LED's brightness in proportion to a changing light level, rather than simply switching the LED fully on or off.
Marking scheme
1 mark: correct description of on/off control (switches fully on/off at a threshold) with a valid example; 1 mark: correct description of continuous control (output varies smoothly/proportionally with input) with a valid example. [2]
Question 8 · Short Answer & Calculation
2 marks
State one advantage of a closed-loop control system, using feedback, over an open-loop control system.
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Worked solution
A closed-loop system continuously measures its actual output and feeds this signal back to be compared with the desired input, automatically adjusting its output to correct for any disturbance or error (for example, maintaining a room at a set temperature despite the door being opened); an open-loop system has no such feedback, so it cannot detect or correct for disturbances once it has been set, making closed-loop control generally more accurate and reliable.
Marking scheme
1 mark: correctly identifies that feedback allows the output to be monitored and errors corrected automatically; 1 mark: correctly contrasts this with an open-loop system's inability to self-correct. [2]
Question 9 · Extended Response (QWC)
9 marks
Discuss the operation of a 4-bit binary counter used as a frequency divider in an electronic control system, explaining how the output frequency at each stage relates to the input clock frequency. Quality of written communication will be assessed in this question.
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Worked solution
A binary counter used as a frequency divider is built from a chain of bistable (flip-flop) stages, typically configured as negative-edge-triggered toggle (divide-by-two) circuits, connected so that the output of each stage becomes the clock input of the next. Each individual bistable changes state (toggles) once for every two pulses received at its clock input, so its output frequency is always exactly half the frequency of the signal driving it. In a 4-bit binary counter, the first stage (Q0) therefore toggles at half the input clock frequency ( \( f_{clock}/2 \) ); the second stage (Q1), clocked from Q0, toggles at \( f_{clock}/4 \); the third stage (Q2) toggles at \( f_{clock}/8 \); and the fourth, most significant stage (Q3) toggles at \( f_{clock}/16 \). In general, a binary counter with n stages divides the input clock frequency by \( 2^n \), so the output taken from the most significant bit of a 4-bit counter is always one-sixteenth of the clock frequency applied to its input. This property makes binary counters useful wherever a lower, precisely related frequency is needed from a higher-frequency clock source, and any of the intermediate stage outputs (Q0-Q2) can also be used if a different division ratio (divide-by-2, -4 or -8) is required within the same circuit.
Marking scheme
Level of Response mark scheme. Level 1 (1-3 marks): basic statement that a counter can divide frequency, with little or no explanation of how; minimal specialist vocabulary; weak spelling/grammar. Level 2 (4-6 marks): satisfactory explanation that each stage toggles at half the frequency of the previous stage, with a generally correct (though possibly incomplete) statement of the overall division ratio; clear use of specialist vocabulary; good spelling/grammar. Level 3 (7-9 marks): comprehensive and accurate explanation covering the divide-by-two action of each bistable stage, correctly stating the frequency at each of the four stages (f/2, f/4, f/8, f/16) and the general \( f_{clock}/2^n \) relationship; highly organised, wide range of accurate technical terms, flawless grammar. [9]
A logic system has three inputs, A, B and C. The output Z is HIGH (1) only for the following input combinations: (A=0,B=0,C=1); (A=0,B=1,C=1); (A=1,B=1,C=1); (A=1,B=1,C=0). For all other combinations of A, B and C, Z is LOW (0). (a) Write out the truth table for this system, listing Z for all eight possible combinations of A, B and C. [3] (b) Use a Karnaugh map to derive the simplified Boolean expression for Z in terms of A, B and C. [3]
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Worked solution
(a) The full truth table (A, B, C, Z) is: 0,0,0 -> 0; 0,0,1 -> 1; 0,1,0 -> 0; 0,1,1 -> 1; 1,0,0 -> 0; 1,0,1 -> 0; 1,1,0 -> 1; 1,1,1 -> 1. (b) Plotting these four 1s on a 3-variable Karnaugh map, the minterms A'B'C and A'BC lie adjacent (differing only in B) and combine to give the term A'.C; the minterms ABC' and ABC lie adjacent (differing only in C) and combine to give the term A.B. No further adjacent-pair combination is possible between these two groups, so the simplified (minimal sum-of-products) expression is \( Z = \overline{A}.C + A.B \). Checking this against every row of the truth table in part (a) confirms it reproduces exactly the same output for all eight input combinations.
Marking scheme
(a) 1 mark for each block of two correctly stated rows (up to 3): all four Z=1 rows correctly identified; the four Z=0 rows correctly identified; all eight rows correctly labelled/ordered. (b) 1 mark: A'B'C and A'BC correctly grouped to give A'.C; 1 mark: ABC' and ABC correctly grouped to give A.B; 1 mark: correct final simplified expression \( Z = \overline{A}.C + A.B \) (accept equivalent notation, e.g. Z = A'C + AB). [6]
A PIC-based control system is required to flash an LED on for 2 seconds and off for 1 second, repeatedly, until a push switch connected to an input pin is pressed, at which point the system should stop flashing and turn the LED off permanently. (a) State, in the correct sequence, the main steps that would appear in a flowchart to achieve this behaviour. [4] (b) State the flowchart symbol (shape) conventionally used to represent a decision, such as checking whether the switch has been pressed. [1] (c) Explain why using a time delay subroutine is useful in this type of program. [1]
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Worked solution
(a) The flowchart would proceed: Start -> set LED output HIGH (LED on) -> call a 2 second time delay subroutine -> set LED output LOW (LED off) -> call a 1 second time delay subroutine -> decision: has the switch been pressed? -> if NO, loop back to setting the LED output HIGH and repeat the cycle; if YES, set the LED output LOW (ensuring it is off) and stop/end the program. (b) A decision is conventionally represented by a diamond (rhombus) shape in a flowchart, with the two possible outcomes (e.g. YES/NO) leading out from two of its points. (c) A time delay subroutine can be written once and then called (reused) at every point in the program where a pause of that length is needed, rather than repeating the same delay code separately each time it is required; this keeps the overall program shorter, easier to read and easier to maintain, since the delay length only needs to be changed in one place if it is ever updated.
Marking scheme
(a) 1 mark each for any four of the following in a logically correct sequence: LED set on; 2 second delay; LED set off; 1 second delay; switch checked as a decision; loop back to on state if not pressed; LED turned off permanently and program stopped if pressed, up to 4. (b) 1 mark: diamond/rhombus. (c) 1 mark: correctly explains reuse of the delay code at multiple points without rewriting it, keeping the program shorter/easier to maintain. [6]
A security system logic circuit must activate an alarm output, Q, only when a door sensor input, D, indicates the door is open AND the system is armed (input S is HIGH), OR when a panic button input, P, is pressed, regardless of the state of D or S. (a) Write a Boolean expression for Q in terms of D, S and P. [2] (b) State the two types of logic gate needed to implement this expression, and explain how they would be connected together. [3] (c) State the output of Q if D=1, S=0 and P=1. [1]
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Worked solution
(a) The alarm Q must be HIGH when D AND S are both HIGH, or when P is HIGH, giving the Boolean expression \( Q = (D.S) + P \). (b) A 2-input AND gate is needed to combine D and S, since both must be true together for that condition to activate the alarm; a 2-input OR gate is needed to combine the result of the AND gate with P, since either condition alone should activate the alarm. The two D and S signals are connected to the two inputs of the AND gate; the output of the AND gate is then connected to one input of the OR gate, with P connected directly to the OR gate's remaining input, and the OR gate's output provides the final alarm signal Q. (c) Substituting D=1, S=0, P=1: D.S = 1 x 0 = 0; Q = 0 + P = 0 + 1 = 1, so Q = 1 (the alarm activates because the panic button has been pressed, regardless of the door/armed state).
Marking scheme
(a) 1 mark: correct AND term for D and S; 1 mark: correctly OR'd with P to give \( Q=(D.S)+P \). (b) 1 mark: AND gate identified for D and S; 1 mark: OR gate identified for combining with P; 1 mark: correct description of the connection (AND output feeds one OR input, P feeds the other). (c) 1 mark: Q = 1, with correct reasoning (D.S=0, P=1, so Q=1). [6]
A microcontroller-based system monitors room temperature using a thermistor input and controls a heater output to maintain the temperature close to a set point of 20 degrees C, switching the heater ON when the temperature falls below 19 degrees C and OFF when it rises above 21 degrees C. (a) Explain why a temperature band (19 degrees C to 21 degrees C) is used, rather than switching the heater at exactly 20 degrees C. [2] (b) State, in the correct sequence, the main steps needed in the flowchart for this control loop. [4]
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Worked solution
(a) If the heater switched exactly at a single 20 degrees C threshold, small fluctuations in temperature around that value (caused by sensor noise or the heater's own effect on the room) would cause the heater to switch on and off very rapidly and repeatedly (a problem known as chattering or hunting); using a wider band (hysteresis), so the heater switches on only below 19 degrees C and off only above 21 degrees C, ensures the heater stays in each state for a sensible period once switched, reducing unnecessary wear on the heater and its switching components. (b) The flowchart proceeds: Start -> read the temperature input from the thermistor -> decision: is the temperature below 19 degrees C? -> if YES, switch the heater output ON; if NO, decision: is the temperature above 21 degrees C? -> if YES, switch the heater output OFF; if NO (temperature is between 19 and 21 degrees C), leave the heater in its current state unchanged -> loop back to read the temperature input again, repeating this cycle continuously.
Marking scheme
(a) 1 mark: correctly identifies the risk of rapid switching/chattering at a single threshold; 1 mark: correctly explains the band prevents this by requiring a wider swing before switching. (b) 1 mark each for any four of: read temperature input; decision comparing to 19 C, switching heater ON if below; decision comparing to 21 C, switching heater OFF if above; heater state left unchanged within the band; loop back to repeat, up to 4. [6]
A PIC output pin is to drive a DC motor rated at 12 V, 250 mA, switched through a single NPN transistor, since the PIC output alone cannot supply enough current to drive the motor directly. (a) State one additional component that must be connected across the motor terminals to protect the transistor, and explain why it is needed. [2] (b) Describe, as a labelled sequence rather than a drawing, how the transistor, motor, protective component and PIC output would be connected. [3] (c) The transistor has a current gain (h_FE) of 100 and must switch the full 250 mA motor current. Calculate the minimum base current required to fully saturate the transistor. [2]
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Worked solution
(a) A flyback (freewheel) diode must be connected across the motor terminals. A motor is an inductive load, and when the transistor switches off, the collapsing magnetic field in the motor's windings generates a large, brief back-EMF voltage spike in the opposite polarity to the supply; without a diode to provide a safe path for this current to dissipate, the spike could exceed the transistor's rated voltage and destroy it. (b) The PIC output pin is connected, via a current-limiting base resistor, to the base of the NPN transistor. The DC motor is connected between the positive supply rail and the transistor's collector. The transistor's emitter is connected to 0 V (ground). The flyback diode is connected directly in parallel across the motor's two terminals, with its cathode towards the positive supply side, so that it remains reverse-biased (non-conducting) during normal operation and only conducts briefly when the motor's back-EMF reverses the polarity across it as the transistor switches off. (c) Minimum base current for saturation: \( I_b = \frac{I_c}{h_{FE}} = \frac{0.250}{100} = 0.0025\ \text{A} = 2.5\ \text{mA} \).
Marking scheme
(a) 1 mark: flyback/freewheel diode identified; 1 mark: correct explanation of back-EMF from the inductive motor load protecting the transistor. (b) 1 mark: PIC output to transistor base (via base resistor); 1 mark: motor between supply and collector, emitter to 0 V; 1 mark: diode correctly placed across the motor with correct polarity described. (c) 1 mark: correct formula/method \( \left(I_b = I_c/h_{FE}\right) \); 1 mark: \( I_b = 2.5\ \text{mA} \). [7]
Question 15 · Major Circuit Design Synthesis
10 marks
Using the blank A3 pro forma provided, design a control system for a garden shed security light that must switch on automatically when it becomes dark AND a passive infrared (PIR) movement sensor detects motion nearby, remaining on for approximately 30 seconds before switching off automatically provided it remains dark. Your design should use an LDR-based darkness sensor, the PIR sensor's digital output, logic gates, a monostable timer, and a suitable output driver for the lamp. Describe: (i) the input stage; (ii) the processing/control stage, including how the two conditions are logically combined and how the 30 second timing is achieved; and (iii) the output stage, including how the lamp is driven.
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Worked solution
(i) Input stage: an LDR is arranged in a voltage-divider circuit with a fixed resistor, and this divider's output feeds a Schmitt trigger, which converts the LDR's gradually changing voltage into a clean, fast-switching digital signal (avoiding false triggering/chatter as the light level crosses the threshold at dusk); this stage outputs a digital HIGH when it is dark. The PIR sensor is a self-contained module with its own internal signal processing, and it provides a digital HIGH output directly on its output pin whenever it detects movement within its field of view, requiring no further conditioning. (ii) Processing/control stage: the digital 'dark' signal from the Schmitt trigger and the digital 'motion detected' signal from the PIR module are each connected to one of the two inputs of a 2-input AND gate; since an AND gate's output only goes HIGH when both of its inputs are HIGH, this ensures the system only proceeds when it is dark AND motion has been detected at the same time. The AND gate's output is connected to the trigger input of a monostable (one-shot) timer circuit, such as a 555 timer configured in monostable mode; the monostable's output pulse width is set by an external resistor-capacitor (RC) timing network chosen to give an output pulse of approximately 30 seconds, and this HIGH output pulse begins as soon as the AND condition is met, regardless of how briefly it was true, with any further trigger while dark simply restarting the 30 second period. (iii) Output stage: because the monostable's output cannot supply the current needed to drive the lamp directly, its output is connected via a base resistor to the base of an NPN transistor, or used to energise a relay coil (with a flyback diode connected across the relay coil to protect the transistor from the coil's back-EMF); the transistor (or the relay's switched contacts) then completes the lamp's supply circuit, switching the lamp on for the full duration of the monostable's 30 second output pulse and switching it off automatically once the pulse ends.
Marking scheme
Up to 3 marks for the input stage: 1 mark LDR/voltage divider correctly identified for darkness sensing; 1 mark Schmitt trigger correctly used to condition the signal; 1 mark PIR module correctly identified as providing a digital motion signal. Up to 4 marks for the processing stage: 1 mark AND gate correctly identified; 1 mark both conditions correctly connected to the AND gate inputs; 1 mark monostable/555 timer correctly identified for the 30 second delay; 1 mark correct explanation that the RC network sets the pulse duration. Up to 3 marks for the output stage: 1 mark transistor or relay correctly identified as the driver; 1 mark flyback diode correctly included for relay protection; 1 mark correct explanation that the lamp is switched for the duration of the monostable pulse. [10]
Question 16 · Major Circuit Design Synthesis
10 marks
Using the blank A3 pro forma provided, design a control system for a conveyor belt item counter that counts items passing a fixed point using an optical slotted switch sensor, and displays the running count (0-9, resetting to 0 after 9) on a single seven-segment display. Describe: (i) how the optical sensor detects each item and produces a suitable pulse for counting; (ii) the counting stage, including the type of counter used and how it is configured to reset after reaching 9; and (iii) the output stage, explaining how the counter's output correctly drives the seven-segment display.
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Worked solution
(i) The slotted optical switch consists of an infrared LED on one side of a slot and a phototransistor on the other side, normally allowing light to pass continuously between them; each time an item passing the fixed point momentarily breaks this light beam as it passes through the slot, the phototransistor's output changes state (e.g. switching from conducting to non-conducting and back), producing one clean voltage pulse for each item that passes; this signal is fed through a Schmitt trigger to sharpen its edges and remove any noise, giving a clean digital pulse suitable for reliably clocking a digital counter. (ii) The conditioned pulses are connected to the clock input of a decade (BCD) counter, a 4-bit binary counter specifically configured to count upward through the binary sequence from 0000 to 1001 (decimal 0 to 9), incrementing its output by one on each qualifying clock edge; unlike a standard 4-bit binary counter, which would continue counting up to 1111 (15), a decade counter automatically resets its output back to 0000 on the clock pulse immediately after it reaches 1001 (9), so the displayed count always cycles correctly through 0-9. (iii) The counter's 4-bit BCD output is connected to the inputs of a BCD-to-seven-segment decoder/driver integrated circuit, which converts each 4-bit binary-coded count into the specific combination of seven-segment outputs (segments a to g) needed to display that digit correctly; the decoder's outputs are then connected directly to the corresponding segments of the seven-segment display (using a decoder/display combination matched as common-anode or common-cathode), so the display always shows the current count value, updating automatically each time the counter increments.
Marking scheme
Up to 3 marks for the sensing stage: 1 mark optical switch/LED and phototransistor correctly identified; 1 mark correct explanation of the beam-break producing a pulse per item; 1 mark Schmitt trigger/signal conditioning correctly included. Up to 4 marks for the counting stage: 1 mark decade/BCD counter correctly identified; 1 mark correct explanation that it counts 0000-1001 (0-9); 1 mark correct explanation that it automatically resets to 0000 after reaching 9; 1 mark correct reference to clocking on each input pulse. Up to 3 marks for the output stage: 1 mark BCD-to-seven-segment decoder/driver correctly identified; 1 mark correct explanation of converting the BCD count to the correct segment pattern; 1 mark correct connection to the seven-segment display described. [10]
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Section B: Mechanical and Pneumatic Control Systems
Answer both Question 3 and Question 4 if choosing the Mechanics/Pneumatics pathway. Complete Question 3(d), Question 4(c), and Question 4(f) on the A3 pro forma answer pages provided.
19 Question · 80 marks
Question 1 · Short Answer & Terminology
2 marks
Define the term 'mechanical advantage' of a machine, and state the formula used to calculate it.
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Worked solution
Mechanical advantage (MA) is the ratio of the load a machine can move (the output force) to the effort force applied to it (the input force): \( MA = \dfrac{Load}{Effort} \). A mechanical advantage greater than 1 means the machine allows a smaller effort to move a larger load.
Marking scheme
1 mark: correctly defines MA as load/effort ratio; 1 mark: correct formula stated \( \left(MA=\frac{Load}{Effort}\right) \). [2]
Question 2 · Short Answer & Terminology
2 marks
Define the term 'velocity ratio' of a machine, and state the formula used to calculate it.
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Worked solution
Velocity ratio (VR) is the ratio of the distance moved by the effort to the distance moved by the load in the same time interval: \( VR = \dfrac{\text{distance moved by effort}}{\text{distance moved by load}} \). It depends only on the geometry of the machine, not on friction or the actual forces involved.
Marking scheme
1 mark: correctly defines VR as the ratio of distances moved by effort and load; 1 mark: correct formula stated. [2]
Question 3 · Short Answer & Terminology
2 marks
State the formula used to calculate the efficiency of a machine, and explain why the efficiency of a real machine is always less than 100%.
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Worked solution
Efficiency is calculated as \( \text{Efficiency} = \frac{MA}{VR} \times 100\% \). A real machine is never 100% efficient because some of the input energy is inevitably lost overcoming friction between its moving parts (and other losses such as the weight of moving components); this lost energy is converted to heat rather than doing useful work on the load, so the mechanical advantage actually achieved is always somewhat less than the velocity ratio would suggest is theoretically possible.
Marking scheme
1 mark: correct formula \( \left(\text{Efficiency}=\frac{MA}{VR}\times100\%\right) \); 1 mark: correct explanation that friction (energy lost as heat) always reduces efficiency below 100%. [2]
Question 4 · Short Answer & Terminology
1 marks
State the SI unit of torque.
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Worked solution
Torque is the turning effect of a force, calculated as force multiplied by the perpendicular distance from the pivot, so its SI unit is the newton metre (N m).
Marking scheme
1 mark: newton metre / N m (accept Nm). [1]
Question 5 · Short Answer & Terminology
1 marks
State the SI unit of power.
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Worked solution
Power is the rate of doing work, measured in watts (W), where one watt equals one joule of work done per second.
Marking scheme
1 mark: watt / W. [1]
Question 6 · Short Answer & Terminology
1 marks
State one function of a proximity sensor in a pneumatic control system.
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Worked solution
A proximity sensor is typically mounted at the end of a cylinder's stroke and detects, without needing physical contact, when the piston has reached that position (fully extended or fully retracted); this feedback signal is used to trigger the next step of a sequential pneumatic circuit.
Marking scheme
1 mark: correctly states it detects cylinder piston position (extended/retracted) to provide feedback for sequencing the next step. [1]
Question 7 · Annotated Sketches & Graphs
3 marks
A cam is designed to produce the following follower motion over one complete revolution (360 degrees): dwell (no movement) for the first 90 degrees; uniform velocity rise of 20 mm from 90 to 180 degrees; dwell at maximum displacement from 180 to 270 degrees; uniform velocity fall of 20 mm back to the start position from 270 to 360 degrees. Describe, in words, how the displacement diagram (a graph of follower displacement against cam angle) for this motion would appear, referring to the shape of the line in each of the four stages.
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Worked solution
From 0 to 90 degrees, the displacement diagram would show a horizontal (flat) line at zero displacement, since the follower does not move at all during the first dwell period. From 90 to 180 degrees, the line would rise as a straight, evenly sloped line from 0 mm up to 20 mm, because uniform velocity motion produces displacement that increases at a constant rate with cam angle, giving a straight line rather than a curve. From 180 to 270 degrees, the line would again be horizontal (flat), this time at the 20 mm displacement level, as the follower dwells at its maximum lift. From 270 to 360 degrees, the line would fall as a straight, evenly sloped line back down from 20 mm to 0 mm, mirroring the rise, since this stage is also uniform velocity motion.
Marking scheme
1 mark: both dwell segments (0-90 deg and 180-270 deg) correctly described as flat/horizontal lines; 1 mark: both uniform velocity segments (90-180 deg and 270-360 deg) correctly described as straight, evenly sloped lines (not curves); 1 mark: correct displacement values referenced (rising from 0 to 20 mm, falling from 20 back to 0 mm). [3]
Question 8 · Annotated Sketches & Graphs
3 marks
Describe, in words, the shape of the displacement diagram for a cam follower undergoing simple harmonic motion (SHM) rise, as opposed to uniform velocity rise, over the same angular range. Explain the key difference in the shape of the graph between the two types of motion.
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Worked solution
For simple harmonic motion (SHM) rise, the displacement diagram forms a smooth, S-shaped (sinusoidal) curve rather than a straight line: the follower starts from rest, accelerates gradually, reaches its maximum velocity at the mid-point of the rise, and then decelerates smoothly back to rest exactly as it reaches the top of the rise. The key difference from uniform velocity motion is in the gradient of the line: uniform velocity motion has a constant gradient throughout the rise, since displacement increases at a fixed rate with cam angle, producing a straight line; SHM has a continuously changing gradient (shallow at the start and end of the rise, steepest at the mid-point), producing a curved line. This means SHM gives the follower a smooth, gradual start and finish to its motion, with no sudden change in velocity at the beginning or end of the rise, unlike uniform velocity motion where the velocity changes abruptly at the start and end of the rise.
Marking scheme
1 mark: correctly describes an S-shaped/sinusoidal curve for SHM; 1 mark: correctly explains gradual acceleration to a maximum at mid-rise, then deceleration to rest (smooth start and finish); 1 mark: correctly contrasts this with the constant gradient/straight line of uniform velocity motion. [3]
Question 9 · Annotated Sketches & Graphs
3 marks
A bell crank lever is used to change the direction of a pull force through 90 degrees, with the pivot positioned so that the input (effort) arm is 120 mm long and the output (load) arm is 40 mm long. (a) Describe, in words, the general arrangement of a bell crank lever. [1] (b) Calculate the velocity ratio of this bell crank lever. [2]
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Worked solution
(a) A bell crank lever is a rigid lever bent through an angle, typically a right angle (90 degrees), at its pivot point, so that its two arms extend away from the pivot in different directions (commonly one arm roughly horizontal and the other roughly vertical); this arrangement allows an input force or motion applied to one arm to be transferred to an output force or motion acting in a different direction at the other arm. (b) Since both arms of the lever rotate through the same angle about the pivot, the ratio of the distances moved by the effort and the load equals the ratio of the arm lengths: \( VR = \dfrac{\text{effort arm length}}{\text{load arm length}} = \dfrac{120}{40} = 3 \).
Marking scheme
(a) 1 mark: correctly describes a lever bent/angled at the pivot, changing the direction of the force/motion. (b) 1 mark: correct method (ratio of arm lengths); 1 mark: VR = 3. [3]
Question 10 · Annotated Sketches & Graphs
4 marks
A simple gear train consists of a driver gear with 20 teeth meshed directly with a driven gear with 60 teeth. (a) State the formula used to calculate the velocity ratio (gear ratio) of a simple gear train in terms of the number of teeth. [1] (b) Calculate the velocity ratio of this gear train. [1] (c) If the driver gear rotates at 300 rpm, calculate the rotational speed of the driven gear. [2]
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Worked solution
(a) For a simple gear train, \( VR = \dfrac{\text{number of teeth on driven gear}}{\text{number of teeth on driver gear}} \). (b) \( VR = \dfrac{60}{20} = 3 \). (c) Output (driven gear) speed = input (driver gear) speed divided by the velocity ratio: \( \text{speed}_{driven} = \dfrac{300}{3} = 100\ \text{rpm} \). This makes sense because the driven gear, having three times as many teeth as the driver, must rotate three times more slowly.
A spanner is used to tighten a nut, with a force of 80 N applied perpendicular to the spanner at a distance of 0.25 m from the centre of the nut. Calculate the torque applied to the nut. State the unit of your answer.
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Worked solution
Torque is calculated as \( T = F \times r \), where F is the force applied and r is the perpendicular distance from the pivot. Substituting: \( T = 80 \times 0.25 = 20\ \text{N m} \).
Marking scheme
1 mark: correct formula used \( \left(T = F \times r\right) \); 1 mark: T = 20; 1 mark: correct unit, N m. [3]
Question 12 · Engineering Calculations
3 marks
A uniform lever is pivoted at its centre. A load of 300 N is placed 0.6 m from the pivot on one side. Calculate the effort required at a distance of 1.2 m from the pivot on the other side to balance the lever.
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Worked solution
By the principle of moments, for the lever to balance: clockwise moment = anticlockwise moment, i.e. \( \text{Load} \times d_{load} = \text{Effort} \times d_{effort} \). Substituting: \( 300 \times 0.6 = \text{Effort} \times 1.2 \), so \( \text{Effort} = \dfrac{300 \times 0.6}{1.2} = \dfrac{180}{1.2} = 150\ \text{N} \).
Marking scheme
1 mark: correct principle of moments used (clockwise moment = anticlockwise moment); 1 mark: correct substitution shown; 1 mark: Effort = 150 N. [3]
Question 13 · Engineering Calculations
3 marks
An electric winch lifts a load of 500 N through a vertical height of 4 m in 10 seconds. (a) Calculate the work done in lifting the load. [1] (b) Calculate the power developed by the winch. [2]
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Worked solution
(a) Work done, \( W = F \times d = 500 \times 4 = 2000\ \text{J} \). (b) Power, \( P = \dfrac{W}{t} = \dfrac{2000}{10} = 200\ \text{W} \).
Marking scheme
(a) 1 mark: W = 2000 J (correct formula and answer). (b) 1 mark: correct formula used \( \left(P=\frac{W}{t}\right) \); 1 mark: P = 200 W. Own figure rule (ECF) applies to part (b) if (a) is incorrect but the method is consistent. [3]
Question 14 · Engineering Calculations
3 marks
A single-acting pneumatic cylinder has a piston diameter of 40 mm and is supplied with air at a gauge pressure of 6 bar. Calculate the maximum force the cylinder can exert on its extension stroke. (Use \( \text{Area} = \pi r^2 \), and 1 bar = 100 kPa = 100 000 Pa.)
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Worked solution
First find the piston area: radius \( r = \dfrac{0.04}{2} = 0.02\ \text{m} \), so \( A = \pi r^2 = \pi \times (0.02)^2 = 1.257 \times 10^{-3}\ \text{m}^2 \). Convert the pressure: \( 6\ \text{bar} = 6 \times 100\,000 = 600\,000\ \text{Pa} \). Force is then \( F = P \times A = 600\,000 \times 1.257 \times 10^{-3} = 754\ \text{N} \) (to 3 s.f.).
Marking scheme
1 mark: piston area correctly calculated \( \left(A \approx 1.257\times10^{-3}\ \text{m}^2\right) \); 1 mark: pressure correctly converted to Pa (600 000 Pa) and force formula used (F = P x A); 1 mark: F = 754 N (accept 750-760 N). [3]
Question 15 · Engineering Calculations
2 marks
A simple pulley system is used to lift a load. The effort end of the rope moves 2.4 m while the load rises 0.6 m. Calculate the velocity ratio of this pulley system.
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Worked solution
\( VR = \dfrac{\text{distance moved by effort}}{\text{distance moved by load}} = \dfrac{2.4}{0.6} = 4 \).
Discuss the operation of a cone clutch, explaining how it transmits torque between two shafts and how the clamping force affects the maximum torque that can be transmitted before slipping occurs. Quality of written communication will be assessed in this question.
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Worked solution
A cone clutch consists of two mating conical friction surfaces: a driving cone, fixed to the input (e.g. engine or motor) shaft, and a driven cone, fixed to the output shaft, which can be moved axially to engage or disengage from the driving cone (typically pressed into engagement by a spring, and released against the spring by a lever or fork mechanism). When the two conical surfaces are pressed together, friction between them allows torque applied to the driving shaft to be transmitted through to the driven shaft, causing both to rotate together. Because the mating surfaces are conical rather than flat, a given axial clamping force produces a larger normal (perpendicular) force pressing the surfaces together than an equivalent flat single-plate clutch would achieve for the same axial force, due to the wedging action created by the cone angle; this is the main practical advantage of a cone clutch, allowing it to transmit more torque for a given actuating force, or to use a smaller actuating force for a given torque requirement. The maximum torque the clutch can transmit before slipping occurs depends on the coefficient of friction between the two surfaces, the normal (clamping) force pressing them together, and the mean radius at which this friction force acts; increasing the axial clamping force (for example, by using a stronger engagement spring) increases the normal force between the surfaces and therefore increases the maximum torque that can be transmitted before slipping. If the torque demanded of the clutch exceeds this maximum, for example under a sudden shock load, the two conical surfaces will slip relative to one another; this limits the torque that reaches the driven shaft but also generates heat and wear at the friction surfaces, so a cone clutch must be sized with an adequate margin of clamping force for its intended application.
Marking scheme
Level of Response mark scheme. Level 1 (1-3 marks): basic description of a cone clutch with little explanation of how torque is transmitted or how clamping force relates to slipping; minimal specialist vocabulary; weak spelling/grammar. Level 2 (4-6 marks): satisfactory explanation of the mating conical surfaces transmitting torque by friction, with some correct reference to clamping force affecting maximum torque before slipping; clear use of specialist vocabulary; good spelling/grammar. Level 3 (7-9 marks): comprehensive and accurate discussion covering the conical friction surfaces and engagement mechanism, the wedging-action advantage of a cone over a flat plate, and a clear, correct explanation of how increasing clamping force increases the maximum torque transmitted before slipping (with consequences of slipping stated); highly organised, wide range of accurate technical terms, flawless grammar. [9]
Question 17 · Major Mechanism / Pneumatic Pro Forma Design
12 marks
Using the blank A3 pro forma provided, design a sequential pneumatic control circuit for two double-acting cylinders, A and B, that must operate in the sequence A+, B+, A-, B- (where + indicates extension and - indicates retraction), returning automatically to the start position ready to begin again when a start push-button is pressed. Your design should use 5/3 directional control valves and proximity sensors for position feedback, and should use an appropriate method (such as a cascade circuit) to avoid signal overlap problems. Describe: (i) how the start signal initiates cylinder A's extension; (ii) how each cylinder's extension/retraction is triggered in sequence using the proximity sensors as feedback; and (iii) how the cascade (or equivalent) method prevents conflicting signals being present at the same valve at the same time.
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Worked solution
(i) Pressing the start push-button sends a pilot air signal to the appropriate pilot input of cylinder A's 5/3 directional control valve, shifting the valve so that it directs supply air to extend cylinder A (A+), beginning the sequence. (ii) A proximity sensor mounted at the fully extended position of cylinder A (sensor a1) detects when A has completed its extension stroke, and its signal pilots cylinder B's 5/3 valve to extend B (B+). A proximity sensor at cylinder B's fully extended position (sensor b1) then detects that B has extended, and its signal pilots cylinder A's valve to retract A (A-). A proximity sensor at cylinder A's fully retracted position (sensor a0) then detects A has retracted, and its signal pilots cylinder B's valve to retract B (B-). Finally, a proximity sensor at cylinder B's fully retracted position (sensor b0) confirms the full sequence has completed and re-arms the start circuit, ready for the next press of the start button. (iii) In a simple direct sequential circuit, a sensor signal (such as a1, which stays active for as long as cylinder A remains extended) can still be physically present at a valve at a later point in the sequence when it is not wanted, causing an unwanted or repeated actuation; a cascade circuit avoids this by dividing the sequence into separate groups (for example, group I covering the A+ step, and group II covering the B+, A- and B- steps), with each group supplied through its own separate air supply line that is switched on or off by cascade (memory) valves as the sequence progresses from one group to the next. Because only the air supply line for the currently active group is pressurised at any one time, a sensor signal belonging to a different group cannot reach and wrongly actuate a valve, ensuring each sensor signal can only take effect at the correct point in the sequence.
Marking scheme
(i) Up to 2 marks: 1 mark start button correctly identified as piloting A's 5/3 valve; 1 mark correctly explains this extends cylinder A. (ii) Up to 6 marks: 1 mark each for correctly identifying a1 triggering B+, b1 triggering A-, a0 triggering B-, and b0 confirming/re-arming the circuit (up to 4); 1 mark for correctly describing proximity sensors as providing non-contact position feedback; 1 mark for the sequence being described in the correct order throughout. (iii) Up to 4 marks: 1 mark correctly identifies the overlap/conflicting signal problem in a direct sequential circuit; 1 mark cascade circuit (or equivalent) correctly identified as the solution; 1 mark correctly explains the sequence is divided into groups on separate air supply lines; 1 mark correctly explains only the active group's line is pressurised, preventing wrongly-timed actuation. [12]
Question 18 · Major Mechanism / Pneumatic Pro Forma Design
12 marks
Using the blank A3 pro forma provided, design a mechanism to convert continuous rotary motion from a small electric motor into reciprocating (back-and-forth) linear motion, to drive a model 'sawing' arm through a stroke of 30 mm. Describe: (i) the mechanism you have chosen and how it converts rotary motion into reciprocating motion; (ii) how the 30 mm stroke length is determined by the dimensions of your chosen mechanism; and (iii) one method of supporting/guiding the moving output to ensure it moves in a straight line.
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Worked solution
(i) A crank and slider mechanism is used to convert the motor's continuous rotary motion into reciprocating linear motion: a short crank arm is fixed eccentrically to the motor's output shaft, and a connecting rod links the outer end of the crank to a slider, which is constrained by guides to move only along a single straight line; as the motor turns the crank through each complete revolution, the connecting rod pushes and pulls the slider back and forth once, producing the required reciprocating 'sawing' motion at the output. (ii) The stroke of the slider (the total distance it travels between its two extreme positions) is equal to twice the radius of the crank, \( \text{stroke} = 2r \), because the slider reaches each of its two extreme positions exactly when the crank arm is aligned with the slider's line of travel, once at each end of the crank's rotation (i.e. across its diameter). To achieve the required 30 mm stroke, the crank radius must therefore be set to \( r = \dfrac{\text{stroke}}{2} = \dfrac{30}{2} = 15\ \text{mm} \). (iii) The slider can be guided using a slot or channel guide, in which a pin or lug fixed to the slider runs inside a straight machined slot, or by supporting the slider on one or more parallel guide rods that pass through holes in the slider body; either method constrains the slider's motion to a single straight line and prevents it twisting or moving sideways as the crank rotates and the connecting rod's angle changes.
Marking scheme
(i) Up to 4 marks: 1 mark crank and slider (or clearly equivalent) mechanism correctly identified; 1 mark crank correctly described as fixed to the motor shaft; 1 mark connecting rod correctly described as linking crank to slider; 1 mark correct explanation that one full crank revolution produces one reciprocating cycle of the slider. (ii) Up to 4 marks: 1 mark correct relationship stated (stroke = 2 x crank radius); 1 mark correct explanation of why (slider at extremes when crank aligns with line of travel); 1 mark correct rearrangement (r = stroke/2); 1 mark correct value, r = 15 mm. (iii) Up to 4 marks: 1 mark valid guiding method identified (slot/channel guide or parallel guide rods); 1 mark correct description of how it is arranged (pin in slot, or rods through slider); 1 mark correct explanation that this constrains motion to a straight line; 1 mark correct explanation that this prevents twisting/sideways movement. [12]
Question 19 · Major Mechanism / Pneumatic Pro Forma Design
11 marks
A small hoist mechanism requires a braking system that holds a suspended load stationary whenever the motor is switched off (including in the event of a power failure), and allows smooth, controlled operation when required. Using the blank A3 pro forma provided, design a suitable braking system. Describe: (i) the type of brake you have selected and justify your choice for this application; (ii) how the brake is applied and released (the actuation method); and (iii) one further safety feature that should be incorporated into your design.
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Worked solution
(i) A spring-applied, electromagnetically-released disc brake is selected for this hoist. A friction pad is pressed against a disc, fixed to the motor or drum shaft, by strong springs whenever the brake's electromagnet is not energised; this makes the brake 'fail-safe', since if electrical power is lost for any reason, the springs automatically clamp the brake on and hold the load in position, rather than allowing it to fall under gravity, which is essential for a hoist given the serious safety hazard an uncontrolled load fall would present. (ii) Whenever the hoist motor is switched on to lift or lower the load under control, an electromagnet built into the brake assembly is energised; this pulls the friction pad back against the spring force, releasing it from the disc and allowing the shaft to rotate freely so the load can be raised or lowered as required. As soon as the motor is switched off, or in the event of a power failure, the electromagnet de-energises immediately, and the springs instantly re-apply the friction pad to the disc, clamping the shaft and holding the suspended load stationary without requiring any further action. (iii) A further safety feature would be an independent secondary (backup) mechanical brake, providing a second means of holding the load if the primary spring-applied brake should ever fail; alternatively, a manual release mechanism that requires a deliberate, sustained action (such as continuously holding a release lever) to disengage the brake, so the brake cannot be released accidentally, is a further appropriate safety feature.
Marking scheme
(i) Up to 4 marks: 1 mark valid brake type identified (e.g. spring-applied/fail-safe disc or drum brake); 1 mark correctly describes the spring-applied clamping arrangement; 1 mark correctly identifies the fail-safe property (holds on loss of power); 1 mark correctly justifies this for a hoist application (prevents uncontrolled load fall). (ii) Up to 4 marks: 1 mark electromagnet correctly identified as the release mechanism; 1 mark correctly explains energising the electromagnet releases the brake against the spring; 1 mark correctly explains de-energising (or power loss) allows the springs to re-apply the brake; 1 mark correct sequencing of this with motor operation. (iii) Up to 3 marks: 1 mark valid further safety feature identified (e.g. secondary/backup brake, manual release requiring sustained action, overload cut-out); 1 mark correct description of how it works; 1 mark correct explanation of the safety benefit it adds. [11]
Section C: Product Design
Answer both Question 5 and Question 6 if choosing the Product Design pathway. Complete Question 5(h) and Question 6(h) on the A3 pro forma answer pages provided.
19 Question · 80 marks
Question 1 · Short Answer & Contextual Application
3 marks
State three of the '6Rs' used in sustainable product design, giving a brief description of each.
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Worked solution
The 6Rs are: Rethink - reconsidering whether and how a product is needed, to reduce its environmental impact from the design stage; Reduce - using less material, energy or packaging in manufacture and use; Reuse - designing a product, or its components, so they can be used again for the same or a different purpose without needing to be reprocessed; Recycle - designing a product so its materials can be recovered and reprocessed into new materials or products at the end of its life; Repair - designing a product so that it can be fixed, extending its useful life rather than it being discarded; and Refuse - avoiding the use of unnecessary or particularly environmentally harmful materials or products altogether.
Marking scheme
1 mark each for any three correctly named and briefly described Rs from: Rethink; Reduce; Reuse; Recycle; Repair; Refuse. [3]
Question 2 · Short Answer & Contextual Application
3 marks
Explain what is meant by a 'life-cycle assessment' (LCA) of a product, and state one stage of a product's life that would be considered in an LCA.
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Worked solution
A life-cycle assessment (LCA) is a systematic evaluation of the environmental impact of a product across the whole of its life, often described as 'cradle to grave': it considers the impact of extracting the raw materials used, manufacturing the product, distributing/transporting it, the product's use by the consumer, and its eventual disposal or recycling at the end of its useful life, allowing designers to identify and target the stages with the greatest environmental impact.
Marking scheme
1 mark: correctly identifies LCA as a systematic evaluation of environmental impact across the product's whole life; 1 mark: correctly conveys the 'cradle to grave' scope (from raw materials to disposal); 1 mark: valid life-cycle stage named. [3]
Question 3 · Short Answer & Contextual Application
2 marks
State two named stages of the product life cycle, in the correct order.
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Worked solution
The product life cycle is generally described in the order: inception (the product is developed/designed), introduction (launched to the market), growth (sales increase as the product becomes established), maturity (sales level off as the market becomes saturated), and decline (sales fall as the product becomes outdated or is replaced).
Marking scheme
1 mark each for any two of inception, introduction, growth, maturity, decline, correctly named and in the correct relative order. [2]
Question 4 · Short Answer & Contextual Application
3 marks
Explain the difference between the life cycle of a 'fad' product and the life cycle of a 'fashion' product.
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Worked solution
A fad product, such as a novelty toy, has a very short and sharp life cycle: it rises to popularity extremely quickly, peaks, and then declines just as rapidly, often disappearing from the market again within a short period. A fashion product, such as a clothing style, has a longer, more gradual life cycle, rising and falling more slowly than a fad, and is often cyclical, sometimes returning to popularity again some years after it first declined, rather than disappearing permanently.
Marking scheme
1 mark: correctly describes a fad's very short, sharp rise-and-fall life cycle; 1 mark: correctly describes a fashion product's longer, more gradual life cycle; 1 mark: correctly notes fashion can be cyclical/recur, unlike a fad. [3]
Question 5 · Short Answer & Contextual Application
2 marks
State two reasons why a manufacturer might choose to redesign an existing product.
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Worked solution
A manufacturer might redesign an existing product to fix an identified fault or problem that has come to light through customer feedback or product testing; to reduce manufacturing or material costs; to update the product's appearance and features so it remains competitive against rival products; to incorporate new technology that has since become available; or to extend the product's life cycle by relaunching an updated version rather than allowing it to decline.
Marking scheme
1 mark each for any two valid, distinct reasons, e.g. fixing a fault; reducing cost; staying competitive/updating appearance; incorporating new technology; extending life cycle/relaunch; improving sustainability. [2]
Question 6 · Short Answer & Contextual Application
3 marks
Explain the benefit of using Computer Numerical Control (CNC) manufacture, rather than manual production, when making a batch of identical plastic components.
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Worked solution
CNC machines follow a stored digital program precisely and repeatably, so every component in a batch is manufactured to exactly the same dimensions and tolerances; this level of accuracy and consistency (repeatability) is very difficult to achieve when producing a batch of components manually, where small variations naturally occur between pieces made by hand. CNC production is also generally faster once the machine has been set up and programmed, and it requires less continuous operator involvement during the actual manufacturing run, which reduces the labour cost per component for a batch or volume production run compared with manual manufacture.
Marking scheme
1 mark: correctly identifies CNC gives high accuracy/consistency (repeatability) between components; 1 mark: correctly explains this is harder to achieve manually; 1 mark: correctly identifies a further benefit (faster once set up/reduced labour cost per component). [3]
Question 7 · Short Answer & Contextual Application
2 marks
Define the term 'Quick Response Manufacturing' (QRM), and state one benefit it offers a manufacturer.
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Worked solution
Quick Response Manufacturing (QRM) is a manufacturing strategy focused on reducing lead times throughout the whole of a company's operations, from when an order is received to when the finished product is delivered to the customer. This allows a manufacturer to respond much more quickly to changing customer demand and individual orders, which in turn reduces the amount of finished stock and raw material inventory that needs to be held in anticipation of future orders.
Marking scheme
1 mark: correctly defines QRM as reducing lead time from order to delivery; 1 mark: valid benefit correctly stated (e.g. faster response to demand, reduced stock/inventory held). [2]
Question 8 · Short Answer & Contextual Application
3 marks
Explain what is meant by 'market pull' in product design, giving an example, and contrast this with 'technology push'.
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Worked solution
Market pull describes a situation where a new product is developed directly in response to an identified need or demand from customers in the market; for example, a manufacturer might develop a smaller, lighter laptop specifically because market research shows customers are increasingly demanding greater portability. This contrasts with technology push, where a new product arises because a new technology, material or manufacturing process has been developed first, and a suitable market or use is then found for it afterwards, rather than the product being developed to meet an already identified customer demand.
Marking scheme
1 mark: correctly defines market pull as product development driven by identified customer demand; 1 mark: valid, correctly matched example given; 1 mark: correctly contrasts with technology push (new technology precedes/drives finding a market). [3]
Question 9 · Short Answer & Contextual Application
3 marks
State the four elements of the marketing mix known as the '4 Ps', and briefly explain what 'Price' covers.
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Worked solution
The 4 Ps of the marketing mix are Product, Price, Place and Promotion. Price covers how a product's selling price is set and the pricing method used to determine it, such as cost-plus pricing, contribution pricing, or pricing based on the perceived value of the product to the customer, and it also covers how a product's pricing strategy may deliberately change as the product moves through the different stages of its life cycle (for example, pricing high on introduction and lowering price as the product matures).
Marking scheme
1 mark: all four Ps correctly named (Product, Price, Place, Promotion); 1 mark: correctly explains Price covers how the selling price/pricing method is determined; 1 mark: correctly notes pricing strategy may vary across the product life cycle (or names a valid pricing method). [3]
Question 10 · Short Answer & Contextual Application
2 marks
Explain what is meant by 'cost-plus' pricing.
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Worked solution
Cost-plus pricing sets a product's selling price by first calculating the total cost of producing it, including materials, labour and overheads, and then adding a fixed profit margin (mark-up) on top of that total cost to arrive at the final selling price.
Marking scheme
1 mark: correctly identifies the total production cost is calculated; 1 mark: correctly identifies a fixed profit margin/mark-up is then added on top. [2]
Question 11 · Short Answer & Contextual Application
3 marks
State two devices used in quality control to ensure that manufactured components are consistently accurate, and explain how one of them is used.
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Worked solution
Devices used in quality control to ensure manufactured components are consistently accurate include jigs, templates, fixtures and gauges/measuring devices. A jig, for example, holds and precisely locates a workpiece and guides a cutting tool, such as a drill, into exactly the correct position; because every workpiece is positioned identically in the jig, every component produced using it has the feature machined in exactly the same place, ensuring accuracy and consistency across the whole production run rather than relying on an operator to measure and position each piece by hand.
Marking scheme
1 mark each for any two valid devices (jigs, templates, fixtures, gauges/measuring devices); 1 mark for a correct explanation of how one of the named devices ensures consistent accuracy. [3]
Question 12 · Short Answer & Contextual Application
2 marks
Explain what is meant by 'inclusive design' when designing a product.
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Worked solution
Inclusive design means designing a product so that it can be used comfortably, safely and effectively by as wide a range of people as reasonably possible, including people of different ages, body sizes, physical abilities and disabilities, rather than designing the product only around the needs of a narrow, 'average' user.
Marking scheme
1 mark: correctly identifies designing for as wide a range of users as possible; 1 mark: correctly references varied user characteristics (e.g. age, size, ability/disability), rather than a single 'average' user. [2]
Question 13 · Short Answer & Contextual Application
3 marks
Explain two ways in which a designer can reduce the environmental impact of a product's packaging.
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Worked solution
A designer can reduce the environmental impact of packaging by minimising the amount of packaging material used to the minimum genuinely needed to protect the product, avoiding unnecessary bulk; by choosing recyclable or biodegradable packaging materials in place of non-recyclable plastics; by designing the packaging shape so that units nest or stack efficiently, reducing wasted space and therefore the number of delivery journeys needed to transport a given number of products; and by avoiding mixed or composite packaging materials, which are difficult to separate and therefore harder to recycle at end of life.
Marking scheme
1 mark each for any two valid, distinct ways, with an explanation, e.g. minimising material used; using recyclable/biodegradable materials; efficient nesting/stacking reducing transport waste; avoiding hard-to-recycle mixed/composite materials, up to 3. [3]
Question 14 · Short Answer & Contextual Application
3 marks
Name a designer or design movement associated with each of the following: (i) a wind-up radio designed to bring communication to areas without a reliable electricity supply; (ii) a bold, colourful postmodern furniture and product design movement of the 1980s; (iii) a company known for minimalist, user-focused product design, associated with products such as the iPhone and iPod.
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Worked solution
(i) Trevor Baylis is associated with the wind-up (clockwork) radio, designed to provide communication in areas without a reliable electricity supply. (ii) The Memphis Group (Memphis) was a bold, colourful, postmodern furniture and product design movement based in Milan in the 1980s, known for rejecting minimalist 'good taste' in favour of playful, decorative forms and clashing colours/patterns. (iii) Apple is known for minimalist, user-focused product design, associated with products such as the iPhone and iPod.
Marking scheme
1 mark each for: (i) Trevor Baylis; (ii) Memphis/Memphis Group; (iii) Apple. [3]
Question 15 · Extended Response (QWC)
9 marks
Discuss the environmental issues that a product designer should consider when designing a new plastic kitchen product, referring to material selection, the 6Rs, and end-of-life disposal. Quality of written communication will be assessed in this question.
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Worked solution
Material selection: the designer should consider using a plastic with recycled content, or a biodegradable/photodegradable plastic where this is genuinely appropriate to the product's expected use and lifespan, weighing this choice against the durability and performance the product actually needs (a biodegradable plastic may not be suitable for a product that must last many years); the embodied energy used to produce the material and how responsibly the raw material is sourced should also be considered, favouring lower-impact materials where performance allows.
The 6Rs: the designer should rethink whether the product genuinely needs to be made from plastic at all, or in its current form; reduce the amount of material used, for example through efficient wall-thickness design that uses no more plastic than structurally necessary; design for reuse where practical, such as making the product refillable rather than single-use; enable repair by using accessible mechanical fixings rather than permanently bonded or welded joints, so a fault does not force the whole product to be discarded; design the product, and its packaging, so its plastic can be recycled at the end of its life, avoiding unnecessary additives, colourants or mixed materials that would complicate recycling; and refuse to include unnecessary single-use plastic components in the design where a suitable alternative exists.
End-of-life disposal: the designer should consider how the product will actually be disposed of once its useful life ends, for example by manufacturing the whole product from a single, easily identifiable plastic type rather than mixed or composite materials that are difficult or uneconomic to separate for recycling, and by marking the product clearly with the correct recycling symbol/information, making it more likely to be correctly sorted and recycled rather than sent to landfill or incineration.
Marking scheme
Level of Response mark scheme. Level 1 (1-3 marks): basic, generic reference to plastic and the environment, with little specific reference to material selection, the 6Rs, or end-of-life disposal; minimal specialist vocabulary; weak spelling/grammar. Level 2 (4-6 marks): satisfactory coverage of at least two of the three named areas (material selection, 6Rs, end-of-life disposal), with some correct, specific detail; clear use of specialist vocabulary; good spelling/grammar. Level 3 (7-9 marks): comprehensive, well-balanced discussion of all three areas, with specific, accurate detail (e.g. recycled/biodegradable material choice and embodied energy; several correctly explained Rs; single-material design and clear recycling marking for disposal); highly organised, wide range of accurate technical terms, flawless grammar. [9]
Question 16 · Graphic Communication & Branding
3 marks
Explain how the choice of text size and style (typography) on a product's packaging can be used to communicate different information to different audiences.
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Worked solution
Large, bold text is typically used for the product name or brand on packaging, creating immediate visual impact and helping the product to be recognised and noticed by a shopper from a distance on a crowded shelf. Smaller, clearer text is used for more detailed information, such as ingredients, usage instructions or safety warnings, which needs to remain legible but does not need to dominate the overall design in the way the brand name does. The particular style or font chosen can also be selected to appeal to and communicate appropriately with the specific target audience for the product; for example, a playful, rounded font might be used for a children's product to appear friendly and fun, while an elegant serif font might be used for a premium product to communicate quality and sophistication.
Marking scheme
1 mark: correctly explains large/bold text is used for the brand name to create impact/recognition; 1 mark: correctly explains smaller, clear text is used for detailed information (ingredients/instructions/warnings) without dominating the design; 1 mark: correctly explains font style is chosen to suit/appeal to the target audience, with a valid example. [3]
Question 17 · Graphic Communication & Branding
3 marks
State three rendering or presentation techniques a designer could use, when producing design proposal drawings by hand, to make a product concept look more realistic and appealing to a client.
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Worked solution
To make a hand-drawn design proposal look more realistic and appealing to a client, a designer could add colour and shading to suggest the material and form of the product; add highlights and shadows to give the drawing a convincing three-dimensional appearance; show the product within a context or setting, such as being held or used, to help the client visualise how it would actually be used; use varying line weights, with a heavier line for the outline and lighter lines for internal detail, to make the drawing clearer and more professional; and add annotation or labels to explain key features or materials that are not otherwise obvious from the drawing alone.
Marking scheme
1 mark each for any three valid rendering/presentation techniques, e.g. colour/shading to show material and form; highlights and shadows for 3D appearance; showing the product in context/use; varied line weights to emphasise outline; annotation/labels explaining features. [3]
A client requires a new range of eco-friendly packaging for a bar of soap, to replace their existing single-use plastic wrapping. Using the blank A3 pro forma provided, design a solution that meets the following criteria: (1) uses recyclable or biodegradable material; (2) uses the minimum material necessary to adequately protect and present the product; (3) clearly communicates the brand name and product information; and (4) is suitable for cost-effective batch production. Describe your design proposal in detail, explaining how it meets each of the four criteria listed, and evaluate one limitation of your chosen solution.
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Worked solution
The proposed packaging is a single-sheet folded card wrap (sleeve) made from uncoated, recycled/recyclable card, printed using water-based or soy-based inks, and closed using a simple fold-and-tuck mechanism formed from the card itself, rather than a separate adhesive label or plastic film overwrap.
(1) Recyclable/biodegradable material: the packaging is made entirely from a single, uncoated recycled card, avoiding plastic film, laminates or plastic-coated finishes that would otherwise prevent it being recycled or composted along with ordinary card/paper waste; using water-based inks (rather than solvent-based inks) further ensures the printed card remains suitable for recycling.
(2) Minimum material: the wrap is cut from a single flat sheet of card, sized as closely as practical to the dimensions of the soap bar itself, with a fold-and-tuck closure formed from the card rather than requiring a separate fastening component; the flat, rectangular die-cut shape is also chosen so that sheets can be nested efficiently when cut from a larger card stock, minimising offcut waste during manufacture.
(3) Communicating brand and product information: the front panel of the wrap is printed with the brand name in large, bold typography for shelf impact and recognition, while a smaller rear or side panel carries the necessary product information (such as ingredients and usage instructions) in smaller, clearly legible text, keeping the two types of information visually distinct without cluttering the main branding.
(4) Suitable for cost-effective batch production: because the design is a single flat card blank, it can be efficiently die-cut and printed while still flat, in large multiples from a single sheet of card stock, before being folded around the soap bar as a final, simple assembly step; this avoids the need for expensive tooling such as injection-moulding equipment, keeping unit costs low for a batch production run.
Limitation: card offers considerably less resistance to moisture than a plastic film wrap, so without a further design provision (such as an internal wax-paper liner, or by only using the wrap for point-of-sale display rather than long-term storage) the soap bar could be affected by damp conditions during storage, transport or display, which is a genuine trade-off against the environmental benefits gained by moving away from plastic packaging.
Marking scheme
Up to 3 marks for each of the four criteria (up to 12 total): 1 mark for a valid feature of the design addressing the criterion; 1 mark for correct detail/justification of how it meets the criterion; 1 mark for a well-developed, specific explanation (not generic). Up to 2 further marks for the evaluation of one limitation: 1 mark for a valid, genuine limitation identified; 1 mark for a correct explanation of its practical consequence. A response that fails to address one or more of the four criteria, or offers no evaluation of a limitation, should be capped accordingly (max 3 marks per unaddressed criterion, max 0 if the limitation is omitted). [14]
A client requires a redesigned handle for a hand-held garden trowel, to be manufactured by injection moulding in a batch production run. Using the blank A3 pro forma provided, design a solution that meets the following criteria: (1) is comfortable and safe to grip for a range of adult hand sizes (inclusive design); (2) is durable and suitable for outdoor/garden use; (3) is suitable for efficient injection-moulded batch manufacture; and (4) incorporates at least one design feature that aids maintenance or repair. Describe your design proposal in detail, explaining how it meets each of the four criteria listed, and evaluate one limitation of your chosen solution.
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Worked solution
The proposed handle has a contoured, ergonomic profile with a slightly flared end, moulded in a durable, UV-stabilised polypropylene for the main body, with a soft-touch thermoplastic elastomer (TPE) over-mould applied in the main grip zone.
(1) Comfortable and safe for a range of hand sizes: the contoured profile follows the general shape of a gripping hand without being so tightly moulded to one specific hand size that it becomes uncomfortable for others; the flared end prevents the hand slipping off the end of the handle during use, and the soft-touch TPE over-mould provides cushioning and additional grip/friction across a range of grip pressures and hand sizes, making the handle inclusive for a wide range of adult users.
(2) Durable and suitable for outdoor use: polypropylene is chosen for the main handle body because it is tough, resists cracking under repeated flexing/impact, and is highly resistant to moisture and most garden chemicals; adding a UV stabiliser to the material prevents the plastic becoming brittle and discoloured from prolonged sunlight exposure, which is important for a tool that will be regularly used and stored outdoors.
(3) Suitable for efficient batch manufacture: the handle is designed with a uniform wall thickness throughout its main body, which reduces the risk of sink marks, warping or uneven cooling during injection moulding; the handle's overall shape is designed with a straightforward, single parting line, allowing it to be produced using a simple two-part mould tool, which keeps tooling costs proportionate and supports efficient, repeatable batch production.
(4) Aids maintenance/repair: the metal trowel blade is attached to the handle using a screwed or bolted fixing, rather than being permanently bonded or over-moulded in place; this allows a blade that becomes worn, bent or damaged to be unscrewed and replaced with a new one, extending the useful life of the handle itself rather than the whole tool having to be discarded when only the blade fails.
Limitation: applying the soft-touch TPE over-mould requires a more complex moulding process than a single-material handle, typically either a two-shot (two-material) injection moulding process or a separate secondary moulding/bonding step; this increases tooling cost and mould complexity compared with a simple single-material handle, which may not be justified if the client's budget is for a very low-cost tool.
Marking scheme
Up to 3 marks for each of the four criteria (up to 12 total): 1 mark for a valid feature of the design addressing the criterion; 1 mark for correct detail/justification of how it meets the criterion; 1 mark for a well-developed, specific explanation (not generic). Up to 2 further marks for the evaluation of one limitation: 1 mark for a valid, genuine limitation identified; 1 mark for a correct explanation of its practical consequence. A response that fails to address one or more of the four criteria, or offers no evaluation of a limitation, should be capped accordingly (max 3 marks per unaddressed criterion, max 0 if the limitation is omitted). [14]
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